Hi, for number 2, say we get $u = f(g(x,y))$ for our general solution when $(x,y) \ne (0,0)$. Then are we just finding a $u(0,0)$ that is equal to the limit of $u = f(g(x,y))$ as $(x,y)$ approaching $(0,0)$ in order to make u continuous at $(0,0)$?

Hi, for number 2, say we get $u = f(g(x,y))$ for our general solution when $(x,y) \ne (0,0)$. Then are we just finding a $u(0,0)$ that is equal to the limit of $u = f(g(x,y))$ as $(x,y)$ approaching $(0,0)$ in order to make u continuous at $(0,0)$?

a) This is a first order linear PDE. We begin by writing equation of characteristic lines: \begin{equation*}\frac{dy}{4y}=\frac{dx}{x}\rightarrow \ln{y}=4\ln{x}+C\end{equation*}\begin{equation*}\Rightarrow \frac{y}{x^4}=C\end{equation*}

This concludes:\begin{equation*}u(x,y)=f(\frac{y}{x^4})\end{equation*}where $f: \mathbb{R} \rightarrow \mathbb{R}$ is an arbitrary function.For $u(x,y)$ to be continuous at $(x,y)=(0,0)$, we should have $$u(0,0)=\lim_{x,y\rightarrow0}{u(x,y)}=0$$For this limit to exist, $\lim_{x,y\rightarrow 0}f(\frac{y}{x^4})$ should exist, meaning $f$ should tend to the limit-value regardless of the path on $x$-$y$ plane in which $x$ and $y$ tend to zero. In particular when we choose $y=Cx^4$ for some C, we get $\lim_{x,y\rightarrow 0}{f(\frac{y}{x^4}})=f(C)$. This being true $\forall C\in \mathbb{R}$ concludes $f$ is a constant function. The only continuous function satisfying PDE is the identically constant function.

b) Analogous to part (a), we start with writing equation of characteristic lines:\begin{equation*}\frac{dy}{-4y}=\frac{dx}{x}\rightarrow \ln{y}=-4\ln{x}+C\end{equation*}\begin{equation*}\Rightarrow yx^4=C\end{equation*}This concludes: \begin{equation*}u(x,y)=f(yx^4)\end{equation*}where $f: \mathbb{R} \rightarrow \mathbb{R}$ is an arbitrary function.This time, $$\lim_{x,y\rightarrow 0}f(yx^4)=\lim_{c \rightarrow 0}f(c)$$For $u(x,y)$ to be continuous we just need to define $u(0,0)=\lim_{c \rightarrow 0}f(c)$. The easiest way to do is of course by getting $f$ continuous at zero and letting $u(0,0)=f(0)$.

The difference between two cases is that in one of them all trajectories have $(0,0)$ as the limit points and in another only those with $x=0$ or $y=0$ (node vs saddle).

Actually since in the saddle case $x^4y=C$ for $C\ne 0$ consists of two disjoint parts (with $x>0$ and with $x<0$), the values of $u$ on these parts are not necessarily equal and $u=f(x^4y)+x|x|^{-1}g(x^4y)$ with $g(0)=0$.